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                Alex Zhang <br />
                Mrs. Compean <br />
                IB Math SL <br />
                November 5, 2012
                <h1 style="text-align:center;">IB Math SL Report: Circles</h1>
                <br />
                Intersecting circles are a baffling subject of geometry that require a hollistic understanding of all the fundamental theorems and postulates of geometry. Similar triangles and proportions are the foundation to this investigation, yet they seemingly have no relation to circles. <br /><br />
              <!--  <div class="alert alert-info">
                    <span class="label label-success">Aim</span><br />The aim of this task is to investigate positions of points in intersecting circles.
                </div>-->
                The following diagram shows a circle \(C_1\) with centre O and radius r, and any point P. <br /><br />
                <div style="text-align:center">
                    <img class="img-polaroid" src="http://i.imagebanana.com/img/5vdsezn0/Selection_015.png" /> 
                </div>
                <br />
                The circle \(C_2\) has centre P and radius OP.  Let A be one of the points of intersection of  \(C_1\) 
                and \(C_2\) 
                .  
                Circle \(C_3\) has centre A, and radius \(r\) .  The point  P' is the intersection of  \(C_3\)  with (OP).  This is 
                shown in the diagram below. <br /> <br />
                <div style="text-align:center">
                    <img src="http://i.imagebanana.com/img/4t7h2mft/Selection_012.png" class="img-polaroid" />
                </div>
              <!--  <h3>Part 1</h3>-->
			  <div>
                What if I let \(r=1\)? I will use an analytic approach to find  OP', when  OP = 2 ,  OP = 3 and  OP = 4 and create a general statement.</div>
                In order to find OP' we must first prove \(\triangle AP'O \sim \triangle PAO\ \)
                <h4>Proof of Similarity</h4>
                <table class="table table-striped table-bordered">
                    <tr>
                        <th>Statement</th>
                        <th>Reason</th>
                    </tr>
                    <tr>
                        <td>
                            \(r = AO,\ C_1\ and\ C_2\text{ have a radius of }r\)
                        </td>
                        <td>
                            Given
                        </td>
                    </tr>
                    <tr>
                        <td>
                            \(C_1 \cong C_2 \)
                        </td>
                        <td>
                            Circles with the same radii are congruent
                        </td>
                    </tr>
                    <tr>
                        <td>\(\overline{AO} = \overline{AP'}\  \)</td>
                        <td> All radii of the same circle are congruent</td>
                    </tr>
                    <tr>
                        <td>\(\triangle AOP' \text{ is isosceles}\ \)</td>
                        <td>Isosceles triangles have two congruent sides</td>
                    </tr>
                    <tr>
                        <td>\(\overline{AP} \cong \overline{OP}\ \)</td>
                        <td> All radii of the same circle are congruent</td>
                    </tr>
                    <tr>
                        <td>\(\triangle PAO \text{ is isosceles}\ \)</td>
                        <td> Isosceles triangles have two congruent sides</td>
                    </tr>
                    <tr>
                        <td>\(\angle AP’O \cong \angle POA \ \)</td>
                        <td> If two sides of a triangle are congruent, the angles opposite them are congruent</td>
                    </tr>
                    <tr>
                        <td>\(\angle POA \cong \angle PAO\ \)</td>
                        <td> If two sides of a triangle are congruent, the angles opposite them are congruent</td>
                    </tr>
                    <tr>
                        <td>\(\angle AP'O \cong \angle PAO\ \)</td>
                        <td> Transitive Property of Congruency</td>
                    </tr>
                    <tr>
                        <td>\(\angle AOP \cong \angle AOP\ \)</td>
                        <td>Reflexive Property of Congruency</td>
                    </tr>
                    <tr>
                        <td>\(\triangle AP'O \sim \triangle PAO\ \)</td>
                        <td> Angle-Angle Similarity</td>
                    </tr>
                </table>
                <!--
                    \[
                    r = AO,\ C_1\ and\ C_2\text{ have a radius of }r |\text{ Given}\\
                    C_1 \cong C_2 \ | \text{ Circles with the same radii are congruent}\\
                    \overline{AO} = \overline{AP'}\  | \text{ All radii of the same circle are congruent}\\
                    \triangle AOP' \text{ is isosceles}\ | \text{ Isosceles triangles have two congruent sides}\\
                    \overline{AP} \cong \overline{OP}\ | \text{ All radii of the same circle are congruent} \\
                    \triangle PAO \text{ is isosceles}\ | \text{ Isosceles triangles have two congruent sides}\\
                    \angle AP’O \cong \angle POA \ | \text{ If two sides of a triangle are congruent, the angles opposite them are congruent} \\
                    \angle POA \cong \angle PAO\ | \text{ If two sides of a triangle are congruent, the angles opposite them are congruent} \\
                    \angle AP'O \cong \angle PAO\ | \text{ Transitive Property of Congruency} \\
                    \angle AOP \cong \angle AOP\ | \text{ Reflexive Property of Congruency }\\
                    \triangle AP'O \sim \triangle PAO\ | \text{ Angle-Angle Similarity}
                    \]
                    -->
                Therefore because the triangles are similar we can set up a proportion of the corresponding parts of the triangles and solve for OP' like so: <br /> <br />
                \(\frac{AO}{OP}=\frac{OP'}{AO}\)<br /><br />
                We can substitute \(AO\) with \(r\) because we are given that fact in the problem. Thus, in all cases \(AO=1\) <br /> <br />
                When \(OP = 2: \)
                \(\frac{1}{2}=\frac{OP'}{1}\);
                \(OP'=\frac{1}{2}\) <br />
                When \(OP=3:\) 
                \(\frac{1}{3}=\frac{OP'}{1}\);
                \(OP'=\frac{1}{3}\) <br />
                When \(OP=4:\)
                \(\frac{1}{4}=\frac{OP'}{1}\); 
                \(OP'=\frac{1}{4}\) <br /> <br />
                <h4>Lengths of OP' when \(r=1\)</h4>
                The following table summarizes my results for various lengths of OP:
                <table class="table table-striped table-bordered">
                    <tr>
                        <th>OP</th>
                        <th>2</th>
                        <th>3</th>
                        <th>4</th>
                    </tr>
                    <tr>
                        <td>OP'</td>
                        <td>\(\frac{1}{2}\)</td>
                        <td>\(\frac{1}{3}\)</td>
                        <td>\(\frac{1}{4}\)</td>
                    </tr>
                    <tr>
                        <td>Graph</td>
                        <td><img class="img-polaroid" style="width:90%;" src="http://i.imagebanana.com/img/3uw9hu3y/Selection_009.png" /></td>
                        <td><img class="img-polaroid" style="width:90%;" src="http://i.imagebanana.com/img/9e4t8vej/Selection_010.png" /></td>
                        <td><img class="img-polaroid" style="width:90%;" src="http://i.imagebanana.com/img/eok3a7sz/Selection_011.png" /></td>
                    </tr>
                </table>
                The computer generated values for what OP' should be match exactly with my values, indicating the proportion is valid even when OP changes in length. A very noticeable pattern is observed – OP' seems to just be the <b>reciprocal</b> of OP, i.e. 1 divided by OP. The formula can be generalized as following: <br /><br />
                <div class="alert alert-success">
                    <span class="label label-inverse"> General Statement Part 1</span> <br />
                    \(OP'=\frac{1}{OP}\)
                </div>
                and looks like this when graphed:<br /><br /><img src="http://i.imagebanana.com/img/dtcexvm0/Selection_014.png" class="img-polaroid" /> <br /> <br />
                <div>
                </div>

                <div>What if I let \(OP=2\) and start varying my radius values to find  \(OP'\), when  \(r=2\), \(r=3\), and \(r=4\) . Moreover, like before I will try to generalize my findings and see if it is consistent with my prior statement. </div>
                Following the earlier proportion we once again use it but this time substitute for the differing \(r\) (aka \(AO\)) cases. <br />
                \(\frac{AO}{OP}=\frac{OP'}{AO}\)<br /><br />
                As stated in the problem, OP is held constant at 2. <br /> <br />
                When \(r=2: \)
                \(\frac{2}{2}=\frac{OP'}{2}\);
                \(OP'=2\) <br />
                When \(r=3:\) 
                \(\frac{3}{2}=\frac{OP'}{3}\);
                \(OP'=\frac{9}{2},4.5\) <br />
                When \(r=4:\)
                \(\frac{4}{2}=\frac{OP'}{4}\); 
                \(OP'=8\) <br /> <br />
                The following table summarizes the results:
                <h4>Lengths of OP' when \(OP=2\)</h4>
                <table class="table table-striped table-bordered">
                    <tr>
                        <th>r</th>
                        <th>OP'</th>
                        <th>Graph</th>
                    </tr>
                    <tr>
                        <td>2</td>
                        <td>2</td>
                        <td><img class="img-polaroid" src="http://i.imagebanana.com/img/6gu1sa21/Selection_018.png" /></td>
                    </tr>
                    <tr>
                        <td>3</td>
                        <td>4.5</td>
                        <td><img class="img-polaroid" src="http://i.imagebanana.com/img/578fsv47/Selection_022.png" /></td>
                    </tr>
                    <tr>
                        <td>4</td>
                        <td>8</td>
                        <td><img class="img-polaroid" src="http://i.imagebanana.com/img/t8sagwh1/Selection_023.png" /></td>
                    </tr>
                </table>
                We can see how the computer generated values in the graphs match the resulting value of OP' indicating the proportion is still valid even with differing values of \(r\). Unlike when \(r\) was held constant and the value of \(OP\) changed, no clear pattern or relationship is apparent between the value \(OP\) and \(OP'\). <br /><br />
                However, as we can see, in order to solve each proportion we must cross multiply; for example, when \(r=2\): \[2\cdot 2=OP'\cdot 2 \\ 4=OP'\cdot 2 \\ OP' = 2 \]
                Written more generally, a pattern emerges: <br /> <br />
                <!--	\[
                    \frac{AO}{OP}=\frac{OP'}{AO} \\
                    \frac{AO}{2}=\frac{OP'}{AO} \text{; OP is always 2 for this part of the question} \\
                    AO \cdot AO = 2 \cdot OP' \\
                    \frac{AO\cdot AO}{2}=OP'\\
                    \frac{AO^2}{2}=OP' \\
                    \frac{r^2}{2}=OP' \text {; We are given } AO = r \text{ so we substitute it in}
                    
                    
                    \]-->
                <table class="table table-striped table-bordered">
                    <tr>
                        <th>Step</th>
                        <th>Comment</th>
                    </tr>
                    <tr>
                        <td>
                            \(\frac{AO}{OP}=\frac{OP'}{AO}\)
                        </td>
                        <td>
                            Initial proportion
                        </td>
                    </tr>
                    <tr>
                        <td>
                            \(\frac{AO}{2}=\frac{OP'}{AO}\)
                        </td>
                        <td>
                            OP is always 2 for this part of the question
                        </td>
                    </tr>
                    <tr>
                        <td>\(AO \cdot AO = 2 \cdot OP' \)</td>
                        <td>Cross Multiply</td>
                    </tr>
                    <tr>
                        <td>\(\frac{AO\cdot AO}{2}=OP' \)</td>
                        <td>Divide both sides by 2</td>
                    </tr>
                    <tr>
                        <td>\(\frac{AO^2}{2}=OP' \)</td>
                        <td>Simplify \(AO \cdot AO\)</td>
                    </tr>
                    <tr>
                        <td>\(\frac{r^2}{2}=OP' \)</td>
                        <td>We are given \(AO = r\) so we substitute it in</td>
                    </tr>
                </table>
                Thus, we come up with our general equation:
                <div class="alert alert-success">
                    <span class="label label-inverse"> General Statement Part 2</span> <br />
                    \(\frac{r^2}{2}=OP' \)
                </div>
                because we notice that every time the value of OP' is derived from multiplying the radius by itself and then dividing it by 2, a quadratic equation is created, graphed here:
                <img class="img-polaroid" src="http://i.imagebanana.com/img/sp8jkvvs/Selection_024.png" /> <br />
                The equation works for all 3 cases:
                \[2^2/2 = 2 \\ 3^2/2=4.5\\4^2/2=8\]
                and is confirmed by our original data from the proportions and also matches the computer computed value from the graphs generated by software as shown in the earlier table. <br /><br />
                This new general equation is not consistent with the earlier general equation; \(\frac{r^2}{2}=OP' \) is only valid in the first part when \(OP=2\), and the earlier general equation \(OP'=\frac{1}{OP}\) is invalid with all values of \(r\).                
                <div>Now I want to use technology to investigate other values of \(r\) and \(OP\) and find the general statement for \(OP'\) .
                    </i>
                </div>
                <!--The following technologies were used:
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                    <li>
                    Geogebra - Graphing software used to set up the relations between the circles</li>
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                    <li>
                    MathJax - Javascript library that parses through LaTeX markup and renders it with HTML + CSS</li>
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                In order to come up with the general statement the following cases are plotted in Geogebra by  manipulating sliders and the computer generated value for OP' is recorded <br />
                <h4>Table depicting values of OP' at arbitrary \(r\) values when \(OP=2\)</h4>
                <table class="table table-striped table-bordered">
                    <tr>
                        <th>r</th>
                        <th>1.5</th>
                        <th>2.5</th>
                        <th>3.25</th>
                    </tr>
                    <tr>
                        <td>OP'</td>
                        <td>1.125</td>
                        <td>3.125</td>
                        <td>5.2813</td>
                    </tr>
                    <tr>
                        <td>Graph</td>
                        <td><img class="img-polaroid" style="width:95%;" src="http://i.imagebanana.com/img/5zba4dnw/Selection_027.png" /></td>
                        <td><img class="img-polaroid" style="width:95%;" src="http://i.imagebanana.com/img/fyb6xxj5/Selection_026.png" /></td>
                        <td><img class="img-polaroid" style="width:95%;" src="http://i.imagebanana.com/img/n7e6mnfn/Selection_025.png" /></td>
                    </tr>
                </table>
                We can see that even for arbitrary values of \(r\) the general statement \(\frac{r^2}{2}=OP' \) calculates the value of OP' correctly <b>as long as the value of \(OP=2\).</b> <br />
                However, when the radius is fixed at 3 and different OP values are used a curious phenomenon is observed:
                <table class="table table-striped table-bordered">
                    <tr>
                        <th>OP</th>
                        <th>3</th>
                        <th>4</th>
                    </tr>
                    <tr>
                        <td>OP'</td>
                        <td>3</td>
                        <td>2.25</td>
                    </tr>
                    <tr>
                        <td>Graph</td>
                        <td><img class="img-polaroid" src="http://i.imagebanana.com/img/feczukri/Selection_028.png" /></td>
                        <td><img class="img-polaroid" src="http://i.imagebanana.com/img/xckpqre2/Selection_029.png" /></td>
                    </tr>
                </table>
                A pattern is noticed and the value of OP' appears to be \(\frac{r^2}{OP}\), which happens to work: 
                \[
                3^3/3 = 3\\
                3^3/4=2.25
                \]
                Looking back at the second general equation \(\frac{r^2}{2}=OP' \) it was assumed the denominator of the fraction was 2 because OP stayed constant; however, that is not the case when examining the first and second parts together holistically. OP changes, and thus \(r^2\) should be divided by OP rather than 2, leading to a new general statement:
                <div class="alert alert-success">
                    <span class="label label-inverse"> General Statement</span> <br />
                    \(\frac{r^2}{OP}=OP' \)
                </div>
                We can also refer back to the original proof, deleting the step that involved substituting OP with 2 because it only applied when OP was held constant at 2, and arrive at the same conclusion.
                <table class="table table-striped table-bordered">
                    <tr>
                        <th>Step</th>
                        <th>Comment</th>
                    </tr>
                    <tr>
                        <td>
                            \(\frac{AO}{OP}=\frac{OP'}{AO}\)
                        </td>
                        <td>
                            Initial proportion
                        </td>
                    </tr>
                    <tr>
                        <td>\(AO \cdot AO = OP \cdot OP' \)</td>
                        <td>Cross Multiply</td>
                    </tr>
                    <tr>
                        <td>\(\frac{AO\cdot AO}{OP}=OP' \)</td>
                        <td>Divide both sides by OP</td>
                    </tr>
                    <tr>
                        <td>\(\frac{AO^2}{OP}=OP' \)</td>
                        <td>Simplify \(AO \cdot AO\)</td>
                    </tr>
                    <tr>
                        <td>\(\frac{r^2}{OP}=OP' \)</td>
                        <td>We are given \(AO = r\) so we substitute it in</td>
                    </tr>
                </table>
                More evidence that dividing by OP is correct is given by the fact that in \(\frac{r^2}{2}=OP' \), 2 is just the length of OP; in \(OP'=1/OP\) the bottom value literally is just OP.
                It becomes clear that the length of OP' is directly proportional to the length of \(r \) while being inversely proportional to the length of \(OP\).                
                <div> In my graphs I noticed that there seemed to be some limits to this statement, so now I will test the validity of my general statement by using different values of OP and r. 
                </div>
                The following table shows various values of OP' under different conditions; these were all generated by the graphing software Geogebra; the equation \(\frac{r^2}{OP}=OP' \) holds true under all these cases.<br /><br />
                <table class="table table-bordered table-striped">
                    <tr>
                        <td></td>
                        <th>
                            OP = 2
                        </th>
                        <th>
                            OP = 3
                        </th>
                        <th>
                            OP = 4
                        </th>
                    </tr>
                    <tr>
                        <th>
                            r = 1
                        </th>
                        <td>\(\frac{1}{2}\)</td>
                        <td>\(\frac{1}{3}\)</td>
                        <td>\(\frac{1}{4}\)</td>
                    </tr>
                    <tr>
                        <th>
                            r = 2
                        </th>
                        <td>2</td>
                        <td>\(\frac{4}{3}\)</td>
                        <td>1</td>
                    </tr>
                    <tr>
                        <th>
                            r = 3
                        </th>
                        <td>\(\frac{9}{2}\)</td>
                        <td>3</td>
                        <td>\(\frac{9}{4}\)</td>
                    </tr>
                    <tr>
                        <th>
                            r = 4
                        </th>
                        <td>8</td>
                        <td>\(\frac{16}{3}\)</td>
                        <td>4</td>
                    </tr>
                </table>
                It should be noted that this formula \(\frac{r^2}{OP}=OP' \) is completely applicable to both Part 1 <b>and</b> Part 2 while previous equations were invalid for some input. For part 1, when \(r=1\), \(r^2\) still equals 1 and it is simply divided by OP, creating the equation that OP' was just the reciprocal of OP, \(OP'=\frac{1}{OP}\):
                \[
                \frac{r^2}{OP}=OP'\\
                \frac{1^2}{OP}=OP'\\
                \frac{1}{OP}=OP'
                \]
                For part 2, OP always equaled 2 so it led to the creation of \(\frac{r^2}{2}=OP' \) from the more general form \(\frac{r^2}{OP}=OP' \) via simple substitution.
                
                <div> There seems to be a scope and limitation of my general statement, which I will discuss. 
                </div>
                There obviously are some limits on the general statement. Firstly, \(r > 0\), \(r \in \mathbb{R}\), \(OP > 0\), \(OP \in \mathbb{R}\) because both are measures which cannot be negative and subsequently must be real numbers. A measure of 0 for either would imply there is no circle.<br />
                Another limit is observed when \(OP=2\) and \(r=4.025\).<br />
                <img class="img-polaroid" src="http://i.imagebanana.com/img/45nhcceg/Selection_030.png" /> <br />
                We can see Point A no longer exists and subsequently \(C_3\) is undefined as well. When \(C_3\) does not exist, OP' can no longer be found, signifying another limit.<br />
                However, when \(OP=2\) and \(r=4\): <br />
                <img class="img-polaroid" src="http://i.imagebanana.com/img/nnfq9h8v/Selection_031.png" /> <br />
                \(C_3\) exists again and the value of OP' can be determined as well. Other values of \(r\) less than 4 that are still within the limitation while \(OP=2\) obviously work as well, as shown by previous graphs.
                <br /> 
                On the other hand, when OP is changed to 3 and \(r=4.025\), \(C_3\) exists again until \(r\) exceeds 6 -- OP' can still be found when \(r=6\), shown in the following diagrams: <br />
                <img class="img-polaroid" src="http://i.imagebanana.com/img/j7htndbv/Selection_032.png" /> <br />
                <img class="img-polaroid" src="http://i.imagebanana.com/img/0z5oobrx/Selection_033.png" /> <br />
                <img class="img-polaroid" src="http://i.imagebanana.com/img/ox4ga3wp/Selection_034.png" />  <br /> <br />
                This occurs because when \(OP=r/2\) the diameter of \(C_2\) happens to be the length of \(r\). Point A, the center of \(C_3\), is defined as the intersection point between \(C_1\) and \(C_2\). And because \(C_2\)'s diameter extends from the center of \(C_1\), the diameter is just the right length for the other end of it to touch \(C_1\), allowing for a point of intersection, aka Point A, to still exist. However, when \(OP&lt;r/2\) there is no longer any point of intersection since the diameter of \(C_2\) is now shorter than the \(r\). Thus, Point A and \(C_3\)would not exist and neither would OP'. <br />
                Therefore, we can conclude there is another limit that \(OP\ge r/2\) in order for OP' to exist and the formula to work.
                <div class="alert alert-success">
                    <span class="label label-inverse">Limitations</span> <br />
                    \(OP\ge r/2\), \(r > 0\), \(r \in \mathbb{R}\), \(OP \in \mathbb{R}\)
                </div>
                <div>In essence, how did I arrive at this general statement?
                </div>
                To recapitulate: finding the general statement came down to using the similar triangles in order to find OP' first and then setting up a proportion of the sides.
                Then, solving the <b>general</b> form of the proportion \(\frac{AO}{OP}=\frac{OP'}{AO}\) from the similar triangles for OP' was vital -- that means just doing a basic algebraic manipulation of the equation to isolate OP' to come up with \(OP'=\frac{AO^2}{OP}\) and then performing a simple substitution to come up with the final form \(OP'=\frac{r^2}{OP}\).
                <h3>Conclusion</h3>
                It is irrefutable that setting up the similar triangle and deriving the proportion from that were the keystone to this problem. Without doing so, the rest of this problem would have been nigh impossible. Being able to visualize the diagrams with different \(r\) and \(OP\) values made a huge difference as well -- the use of Geogebra to graph the circles played a critical role in the completion of this task.
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